# Homework Help: Infinite series - Inverse trigonometry

1. Aug 18, 2013

### Saitama

1. The problem statement, all variables and given/known data
The sum of the infinite terms of the series
$$\text{arccot}\left(1^2+\frac{3}{4}\right)+\text{arccot}\left(2^2+\frac{3}{4}\right)+\text{arccot}\left(3^2+\frac{3}{4}\right)+............$$
is equal to
A)arctan(1)
B)arctan(2)
C)arctan(3)
D)arctan(4)

Ans: B

2. Relevant equations

3. The attempt at a solution
The given series can be written as
$$\sum_{r=1}^{\infty} \text{arccot}\left(r^2+\frac{3}{4}\right)$$
Simplifying the term inside the summation, I rewrite it as
$$\arctan\left(\frac{4}{4r^2+3}\right)$$
I don't see how to proceed further. I am thinking of converting the above in the form $\arctan\left(\frac{x-y}{1+xy}\right)$, rewrite it in the form $\arctan(x)-\arctan(y)$ and perform the summation but I can't see a way to do this.

Any help is appreciated. Thanks!

Last edited: Aug 18, 2013
2. Aug 18, 2013

### vela

Staff Emeritus
I haven't worked it out, but I think you're on the right track. Try rewriting it as
$$r^2 + \frac{3}{4} = \left(r^2 - \frac{1}{4}\right) + 1 = \left(r+\frac{1}{2}\right)\left(r-\frac{1}{2}\right) + 1$$

3. Aug 18, 2013

### Saitama

Wow. That's great. Thanks a lot vela!

I did try rewriting 3/4 as 1-1/4 but then I didn't notice that it could simplified to (r-1/2)(r+1/2).

4. Aug 19, 2013

### verty

This is one of the most sadistic questions I've seen. Where are you getting these questions?

5. Aug 19, 2013

### Saitama

From practice sheets. Do you want me to send the links to those practice sheets by PM? I don't know if its ok to post them here.

6. Aug 19, 2013

### verty

A PM will be fine, thanks. I haven't seen these type questions before, is all.